What is the mass ratio of ethylene glycol (C$$_2$$H$$_6$$O$$_2$$, molar mass = 62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molar aqueous solution?

For the first case, let the mass of ethylene glycol be (w_1) g. Since the solution is (0.25) molal,

$$0.25=\frac{w_1/62}{(500-w_1)/1000}.$$

On solving,

$$0.25=\frac{1000w_1}{62(500-w_1)},$$

$$15.5(500-w_1)=1000w_1,$$

$$7750-15.5w_1=1000w_1,$$

$$7750=1015.5w_1,$$

$$w_1\approx7.63\text{ g}.$$

For the second case, the solution is (0.25) M and its volume is (250) mL (=0.250) L.

The number of moles of ethylene glycol is

$$0.25\times0.250=0.0625\text{ mol}.$$

Hence, the required mass is

$$w_2=0.0625\times62=3.875\text{ g}.$$

Therefore, the required mass ratio is

$$\frac{w_1}{w_2}=\frac{7.63}{3.875}\approx1.97\approx2.$$

Hence, the mass ratio is

$$\boxed{2:1}.$$